(a) Explanation Given information: The speed of the block slide ( v ) is 8 ft / s . The height from horizontal surface to the block ( h ) is 3 ft . The acceleration due to gravity ( g ) is 32.2 ft / s 2 . Neglect friction and air resistance. Calculation: Show the free body diagram of the block with the forces acting as in Figure (1). The normal force ( N ) will be zero when the block leaves the surface at C . Determine the vertical force by considering the vertical equilibrium condition. N + m g cos θ = m a n Substitute 0 for N and v C 2 h for a n . 0 + m g cos θ = m ( v C 2 h ) g cos θ = v C 2 h v C 2 = g h cos θ v C 2 = g y (1) Take h cos θ for y . Calculate the initial kinetic energy ( T B ) using the formula: T B = 1 2 m v 2 Substitute 8 ft / s for v . T B = 1 2 m ( 8 ) 2 = 32 m The expression for calculating the final kinetic energy ( T C ) as follows; T C = 1 2 m v C 2 Calculate the work done ( U B − C ) using the formula: ( U B − C ) = W × ( h − y C ) Substitute m g for W . ( U B − C ) = m g ( h − y C ) Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement. The expression for the principle of work and energy for the earth is as follows; T B + U B − C = T C Substitute 32 m for T B , m g ( h − y C ) for ( U B − C ) , and 1 2 m v C 2 for T C (b) To determine Find the distance ( x ) at which the block will hit the ground.

12th Edition

ISBN: 9781260804638

Author: BEER

Publisher: MCG

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Chapter 13.1, Problem 13.45P

(a)

To determine

Find the angle (θ) at which the block will leave the cylindrical surface BCD.

(b)

To determine

Find the distance (x) at which the block will hit the ground.

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Chapter 13.1, Problem 13.44P

Chapter 13.1, Problem 13.46P

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Find the angle ( θ ) at which the block will leave the cylindrical surface BCD .

Solution Summary: The author shows the free body diagram of the block with the forces acting as in Figure (1).